Overview
Comment: | 202409 1st star |
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Downloads: | Tarball | ZIP archive | SQL archive |
Timelines: | family | ancestors | descendants | both | trunk |
Files: | files | file ages | folders |
SHA3-256: |
868207236374b9594ad435b7a82432c0 |
User & Date: | nnz on 2024-12-09 11:50:44 |
Other Links: | manifest | tags |
Context
2024-12-09
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12:24 | changed empty values to work with part two check-in: c93635c614 user: nnz tags: trunk | |
11:50 | 202409 1st star check-in: 8682072363 user: nnz tags: trunk | |
2024-12-08
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17:11 | 202408 2nd star check-in: 9395df6993 user: nnz tags: trunk | |
Changes
Modified aoc2024.c from [05505307b8] to [3ba7100c1a].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #include <ctype.h> #include <stdbool.h> #include <stddef.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include "aocdailies.h" #include "aocutils.h" /* === aoc202408 ======================================================= Oh! another square of text! Idea: for all points p with an antenna find all points q>p with a corresponding frequency. For each such pair calculate the 2 antinodes and add the resulting points (if not there already) to an array. The answer to Part One is the number of elements in the array | > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 | #include <ctype.h> #include <stdbool.h> #include <stddef.h> #include <stdio.h> #include <stdlib.h> #include <string.h> #include "aocdailies.h" #include "aocutils.h" /* === aoc202409 ======================================================= WOW! input consists of a 20,000 long string of numbers! ===================================================================== */ void aoc202409(char *data, size_t len) { (void)len; // unused argument int *disk = malloc(512 * sizeof *disk); size_t rdisk = 512; size_t ndisk = 0; int fileid = 0; while (*data) { for (int k = 0; k < *data - '0'; k++) { if (ndisk == rdisk) { // assume it works rdisk = (13*rdisk)/8; disk = realloc(disk, rdisk * sizeof *disk); } disk[ndisk++] = fileid; } fileid++; data++; if (*data) { for (int k = 0; k < *data - '0'; k++) { if (ndisk == rdisk) { // assume it works rdisk = (13*rdisk)/8; disk = realloc(disk, rdisk * sizeof *disk); } disk[ndisk++] = -1; } data++; } } int *left = disk, *right = disk + ndisk; for (;;) { while (*left != -1) left++; if (left >= right) break; while (right[-1] == -1) right--; // swap *left and right[-1] int tmp = *--right; *right = *left; *left++ = tmp; } unsigned long long chksum = 0; size_t index = 0; while (disk[index] >= 0) { chksum += (size_t)disk[index] * index; index++; } printf("%llu\n", chksum); free(disk); } /* === aoc202408 ======================================================= Oh! another square of text! Idea: for all points p with an antenna find all points q>p with a corresponding frequency. For each such pair calculate the 2 antinodes and add the resulting points (if not there already) to an array. The answer to Part One is the number of elements in the array |
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Modified aocdailies.c from [fb8ca294f1] to [53d57bd116].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #include <stddef.h> #include "aocdailies.h" aocfunc *aocselect(unsigned y, unsigned d) { aocfunc *p; switch (y * 100 + d) { default: p = NULL; break; // YYYYdd ==> aocYYYYdd case 202408: p = aoc202408; break; case 202407: p = aoc202407; break; case 202406: p = aoc202406; break; case 202405: p = aoc202405; break; case 202404: p = aoc202404; break; case 202403: p = aoc202403; break; case 202402: p = aoc202402; break; | > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | #include <stddef.h> #include "aocdailies.h" aocfunc *aocselect(unsigned y, unsigned d) { aocfunc *p; switch (y * 100 + d) { default: p = NULL; break; // YYYYdd ==> aocYYYYdd case 202409: p = aoc202409; break; case 202408: p = aoc202408; break; case 202407: p = aoc202407; break; case 202406: p = aoc202406; break; case 202405: p = aoc202405; break; case 202404: p = aoc202404; break; case 202403: p = aoc202403; break; case 202402: p = aoc202402; break; |
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Modified aocdailies.h from [114faad3c2] to [e29d12fccf].
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 | #ifndef AOCDAILIES_H_INCLUDED #define AOCDAILIES_H_INCLUDED #include <stddef.h> typedef void aocfunc(char *, size_t); aocfunc *aocselect(unsigned, unsigned); aocfunc aoc202408; aocfunc aoc202407; aocfunc aoc202406; aocfunc aoc202405; aocfunc aoc202404; aocfunc aoc202403; aocfunc aoc202402; | > | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | #ifndef AOCDAILIES_H_INCLUDED #define AOCDAILIES_H_INCLUDED #include <stddef.h> typedef void aocfunc(char *, size_t); aocfunc *aocselect(unsigned, unsigned); aocfunc aoc202409; aocfunc aoc202408; aocfunc aoc202407; aocfunc aoc202406; aocfunc aoc202405; aocfunc aoc202404; aocfunc aoc202403; aocfunc aoc202402; |
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